Here is the complete question
An aqueous solution is saturated with both a solid and a gas at 5 °C. What is likely to happen if the solution is heated to 85 °C ?
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a.) Some gas will bubble out of solution and more solid will dissolve.
b.) Some gas will bubble out of the solution and some solid will precipitate out of the solution.
c.) Some solid will precipitate out of solution.
d.) More gas will dissolve and more of the solid will dissolve.
Answer:
a.) Some gas will bubble out of solution and more solid will dissolve.
Explanation:
Temperature increase usually increases the dissolution of solids in liquids. From the question; Some gas will be bubble out of the solution as the temperature is being increased to 85 °C because that aqueous solution is saturated(i.e equal amount of solute and solvent in the solution) with both solid and gas at 5 °C, but when the solution is heated to 85 °C, the solution becomes supersaturated( i.e the solute is now more at the given temperature than the solvent).
Answer:
As the temperature increases, the solubility of the solid increases and the solubility of the gas decreases. When the solution that is saturated between a solid and a gas at 5 ° C and heated to 85 ° C, the gas comes out of the solution first.
Explanation:
A saturated solution is one that has the maximum amount of solute that is dissolved. An unsaturated solution is one that has a low amount of solute compared to the saturated solution. According to Henry's law, the solubility of a gas at a specific temperature is directly proportional to its partial pressure, that is
C ∝ p
C = kp
Where
p is the partial pressure
k ia a proportionality constant
C is the concentration of the gas
CH3CH2OH(l) 3O2(g) Classify each chemical reaction: reaction type of reaction (check all that apply) combination precipitation single replacement combustion double replacement acid-base decomposition combination precipitation single replacement combustion double replacement acid-base decomposition combination precipitation single replacement combustion double replacement acid-base decomposition combination precipitation single replacement combustion double replacement acid-base decomposition
Answer:
Reaction I is a COMBINATION REACTION - A reaction that involves the mixing of two or more elements to form a single product.
Reaction II is COMBUSTION REACTION - A reaction that involves Oxygen to produce carbon(iv)oxide and water vapor
Reaction III is DOUBLE DISPLACEMENT REACTION - A reaction that involves the exchange of radicals.
Reaction IV is a COMBUSTION REACTION
Explanation:
Reaction I is a COMBINATION REACTION - A reaction that involves the mixing of two or more elements to form a single product.
Reaction II is COMBUSTION REACTION - A reaction that involves Oxygen to produce carbon(iv)oxide and water vapor
Reaction III is DOUBLE DISPLACEMENT REACTION - A reaction that involves the exchange of radicals.
Reaction IV is a COMBUSTION REACTION
Attached is the reactions I - 1V
The given chemical equation CH3CH2OH(l) + 3O2(g) represents a combustion reaction where CH3CH2OH reacts with oxygen to produce carbon dioxide and water.
Explanation:Based on the given chemical equation, CH3CH2OH(l) + 3O2(g), the reaction is a combustion reaction. Combustion reactions involve the rapid combination of a fuel (in this case, CH3CH2OH) with oxygen (O2) to produce heat, light, and new products. In a combustion reaction, a fuel is oxidized and reacts with oxygen to form carbon dioxide and water.
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After doing an experiment, a chemist determines the Rf value of a compound to be 4. He also notes that the solvent travelled 4 cm on the plate. What can you conclude about this experiment
Answer:
We can conclude that the Rf of that compound has a ratio of 4. It means that the solute has a ratio value of 4 times than that of solvent. As we can see that it has traveled 4 cm , this data is useful in determination of the compound in a mixture when compared with Rf values of other compounds.
A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of cold water at 0∘C (cwater=4186J/kg⋅∘C) and sealed. At equilibrium, the temperature of the water and block are measured to be 20∘C.
0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of cold water at 0∘C (cwater=4186J/kg⋅∘C) and sealed. At equilibrium, the temperature of the water and block are measured to be 20∘C.
If the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water and block?
A. less than 20∘C
B. 20∘C
C. greater than 20∘C
Answer:
Option C is correct
Explanation:
Increase in the mass of aluminium block would increase the heat capacity of block isothermaly before immersed in water at 0° so heat available for transfer is higher so equilibrium temperature of system would increase.
Given that D(H-H) and D(F-F) in H2 and F2 are 436 and 158kJ mol-1, estimate the bond dissociation enthalpy of H-F using a simple additivity rule. Compare the answer with the experimental value of 570kJ mol-1
Explanation:
Equation of the reaction:
H2(g) + F2(g) --> 2HF(aq)
1/2H2(g) + 1/2F2(g) --> HF(aq)
D(H-H) in H2 = 436 kJ/mol
D(F-F) in F2 = 158kJ/mol
ΔH bond breakage (dissociation):
1/2 mol H-H bonds = (1/2 X 436) kJ
= 218 kJ
1/2 mol F-F bonds = (1/2 X 158) kJ = = 80 kJ
Total = 218 + 80 = 298 kJ
ΔH bond formation:
1 mol H-F bonds = - 570 kJ
= DHreactant - DHproduct
ΔH°f = 298 kJ + -570 kJ = -272 kJ
(a) The first step in ozone formation in the upper atmosphere occurs when oxygen molecules absorb UV radiation of wavelengths ≤ 242 nm. Calculate the frequency and energy of the least energetic of these photons. (b) Ozone absorbs light having wavelengths of 2200 to 2900 Å, thus protecting organisms on Earth’s surface from this high-energy UV radiation. What are the frequency and energy of the most energetic of these photons?
Answer:
a) f = (1.24 × 10^15) Hz and E = (8.214 × 10^-19) J
b) f = (1.36 × 10^15) Hz; E = (9.035 × 10^-19) J
Explanation:
a) The least energetic photons have the highest wavelength. That is, the wavelength of the least energetic photons is equal to the upperlimit of the wavelength inequality given.
λ = 242nm = 2.42 × 10⁻7 m
v = fλ; f = v/λ; v = 3×10^8 m/s
f = (3×10^8)/(2.42×10^-7)
f = (1.24 × 10^15) Hz
E = hf; h = planck's constant = (6.62607004 × 10^-34) Js
E = 6.626 × 10^-34 × 1.24 × 10^15
E = (8.214 × 10^-19) J
b) The photons with the least wavelength in the range provided are the most energetic ones.
λ = (2200 × 10^-10) m = (2.2 × 10^-7) m
v = fλ; f = v/λ; v = 3×10^8 m/s
f = (3×10^8)/(2.2×10^-7)
f = (1.36 × 10^15) Hz
E = hf; h = planck's constant = (6.62607004 × 10^-34) Js
E = 6.626 × 10^-34 × 1.36 × 10^15
E = (9.035 × 10^-19) J
QED!
In ozone formation, least energetic photons having wavelength 242 nm have frequency ≈ 1.24 x 10^15 Hz and energy ≈ 8.2 x 10^-19 J. The most energetic photons that ozone absorbs, with a wavelength of 2200 Å, have a frequency of ≈ 1.36 x 10^15 Hz and energy of ≈ 9.02 x 10^-19 Joules.
Explanation:(a) The frequency (ν) of a photon is given by the formula: ν = c / λ, where c is the speed of light (3.00 x 10^8 m/s) and λ is wavelength. For the least energetic photons (wavelength of 242 nm), we would convert the wavelength to meters (242 nm = 242 x 10^-9 m). Applying the formula: ν = 3.00 x 10^8 m/s / 242 x 10^-9 m, we get ν ≈ 1.24 x 10^15 Hz.
The energy (E) of a photon is given by the formula: E = hν, where h is Planck’s constant (6.63 x 10^-34 Js). So, E = 6.63 x 10^-34 Js x 1.24 x 10^15 Hz, which gives us E ≈ 8.2 x 10^-19 Joules.
(b) For the most energetic of these photons, they have the shortest wavelength (2200 Å = 2200 x 10^-10 m). Using similar calculations as above: ν ≈ 1.36 x 10^15 Hz and E ≈ 9.02 x 10^-19 Joules.
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How are measurements of paramagnetism used to support electron configurations derived spectroscopically? Use Cu(I) and Cu(II) chlorides as examples.
Answer:
Paramagnetism is dependent on the unpaired electron in the last orbital . In this regard, Cu(I) chloride is paramagnetic whereas Cu(II) chloride is not.
Explanation:
Paramagnetism is the property of materials/components which makes them attracted them weekly to the magnetic field.
It is related to electronic configuration, such that it depends on the unpaired electron in the last orbital possess the property.
On basis of this property, Cu(I) chloride is paramagnetic while Cu(II) chloride is non paramagnetic. This is because Cu(I) chloride contains an unpaired electron in the last orbital whereas Cu(II) chloride does not have any unpaired electron.
Answer:
Explanation:
Paramagnetism is a type of magnetism whereby materials are weakly attracted to an externally applied magnetic field and then form internal, induced magnetic fields in the direction of the applied magnetic field. They are attracted to magnetic fields and have magnetic moment induced by the applied field is linear in the field strength. Paramagnetic materials include elements such as Oxygen,
Aluminium etc. and maybe some compounds like FeO etc.
Paramagnetism occurs due to the presence of unpaired electrons in an atom, so atoms with incompletely filled atomic orbitals are paramagnetic, there are exceptions such as copper exist and this is due to their spin, unpaired electrons have a magnetic dipole moment and act like tiny magnets. They have a magnetic permeability slightly greater than 1. External magnetic field causes the electrons spin to align parallel to the field hence, causing a net attraction. Paramagnetic materials include aluminium, oxygen, titanium, and iron oxide (FeO).
From the example,
Cu(I) and Cu(II)
Electronic configuration
Cu(I) - [Ar] 3d10
Cu(II) - [Ar] 3d9
[Ar] - 1s2 2s2 2p6 3s2 3p6 4s2
Therefore, Cu(I) is Paramagnetic while Cu(II) is not Paramagnetic (diamagnetic).
1 message Which process results in the increase in entropy of the universe? the cooling of a hot cup of coffee in room temperature air the melting of snow above 0 °C all of the above the evaporation of water from a desk at room temperature
Answer:
Melting of snow
Evaporation of water from desk
Explanation:
Processes that increase the entropy of the universe are those processes that have an increased disorderliness. We should note that there are three principal states of matter which are the liquid, gas and solid. The gaseous state is the most disorderly while the solid is the least disorderly.
Now. We can see that the cooling of a hot cup of coffee is a process that needs or leads to a loss in temperature which obviously decreases disorderliness of the universe.
The melting of snow however is a process that leads to an increase in the disorderliness of the universe. It entails moving from the solid state to the liquid state. It tends to move to a more disordered state indicating an increase in the entropy of the universe.
The evaporation of water from the desk is quite similar to that above. Hence since we are moving from the liquid to the gaseous state via evaporation, we can state that the entropy of the universe has increased since we have moved from a state with a lesser degree of disorderliness to a state that is more disordered I.e from liquid to gaseous state.
The processes that result in the increase in entropy of the universe would be the melting of snow above 0 °C and the evaporation of water from a desk at room temperature.
Entropy refers to the degree of disorderliness of a system. The degree of disorderliness increases when the temperature of a system increases or when there is a phase change in the order solid > liquid > gas.
Thus, when snow melts, a phase change occurs from solid to liquid leading to an increase in entropy. Also. when water evaporates from a desk, it changes phase from liquid to gas and thereby leads to an increase in entropy.
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Lithium has two naturally occurring isotopes, 6Li and 7Li . The atomic weight of lithium is 6.941. Which of the following statements concerning the relative abundance of each isotope is correct? A) The abundance of 7Li is greater than 6Li. B) The abundance of 7Li is less than 6Li. C) The abundance of 6Li is equal to the abundance of 7Li. D) Not enough data is provided to determine the correct answer. E) Based on the atomic mass, only 7Li occurs naturally.
Answer: The abundance of Li-7 isotope is higher as compared to Li-6.
Explanation:
Average atomic mass is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]
We are given:
Two isotopes of lithium :
Li-6 and Li-7
Average atomic mass of lithium= 6.941
As, the average atomic mass of lithium is closer to the mass of isotope Li-7. This means that the relative abundance of Li-7 is higher as compared to Li-6.
Percentage abundance of Li-7> Percentage abundance of Li-6 isotope
The atomic weight of lithium is closer to the mass number of 7Li, indicating that 7Li is more abundant than 6Li in nature. Thus, the correct answer is A) The abundance of 7Li is greater than 6Li.
Explanation:The atomic weight of lithium, being 6.941, is closer to 7 than 6. Consequently, this indicates that most naturally occurring lithium is of the heavier 7Li isotope. So, in terms of relative abundance, 7Li is indeed more prevalent than 6Li. This means the correct answer would be option A) The abundance of 7Li is greater than 6Li. Therefore, based on the atomic weight of lithium, we can conclude that the abundance of 7Li is greater than 6Li.
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(a) which of these gases would you expect to have the largest van der waals constant a? H2, HF, F2******** (b) which of these gases would you expect to have the largest van der waals constant b? H2, HCL, CL2
In the given gases, HF should have the largest van der Waals constant a because of its stronger intermolecular forces, and Cl2 should have the largest van der Waals constant b due to its larger molecular size.
Explanation:The van der Waals constants (a) and (b) are indicative of the strength of the forces between molecules, and the physical size of the molecules in a given gas, respectively.
(a) In the selection of H2, HF, F2, we would expect the molecule with the strongest intermolecular forces to have the largest van der Waals constant a. That would be HF because when comparing these gases, HF has permanent dipole-dipole interaction which is stronger than the London dispersion forces in H2 and F2.
(b) For the gases H2, HCl, Cl2, we would expect the molecule with the largest physical size to have the largest van der Waals constant b. In this case, Cl2 is the largest molecule and thus would have the largest van der Waals constant b under normal conditions.
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Draw the four structures of the compounds with molecular formula C5H10O that contain a carbon-carbon double bond, an unbranched carbon chain, and a hydroxyl group one carbon from the end of the chain.
Explanation:
Molecular formula of the compound =
The carbon chain in the molecule will look like :
1. [tex]H_2C=CH-CH_2-CH_2-CH_2-OH[/tex]
2. [tex]H_3C-CH=CH-CH_2-CH_2-OH[/tex]
3. [tex]H_3C-CH-CH=CH-CH_2-OH[/tex]
4. [tex]H_3C-CH_2-CH_2-CH=CH-OH[/tex]
No branching is present, so that means the valency of the carbon will be fulfilled by 10 hydrogen atoms and 1 oxygen atom
Suppose you want to test the results of a transformation by growing Escherichia coli cells in LB medium containing ampicillin as the antibiotic for selection. Ampicillin at a concentration of 100 µg/mL will kill cells that do not contain an ampicillin resistance gene, but will allow the growth of cells that have been transformed with this gene. The concentrated stock of ampicillin is 100 mg/mL. How many microliters of the ampicillin stock should you add to 50 mL of LB for a bacterial culture?
Answer:
50.0 μL
Explanation:
When a dilution is done, the mass of the solute (in this case the ampicillin) remains constant, following the Lavoiser's law that the mass is conserved. The mass is the concentration (C) multiplied by the volume (V), so if 1 is the stock solution, and 2 is the bacterial culture after the addition of the antibiotic:
m1 = m2
C1*V1 = C2*V2
C1 = 100 mg/mL = 100000 μg/mL (1 mg = 1,000μg)
C2 = 100 μg/mL
V2 = 50 mL + V1 = 50000μL + V1 (V1 in μL)
100000*V1 = 100*(50000 + V1)
1000V1 = 50000 + V1
999V1 = 50000
V1 = 50.0 μL
An aqueous solution is 36% by mass silver nitrate, AgNO3, and has a density of 1.41 g/mL. What is the molality of silver nitrate in the solution?
To find the molality of the silver nitrate solution, we can use the mass percent and density of the solution. The molality is found to be 2.99 mol/kg.
Explanation:To find the molality of the silver nitrate solution, we need to first determine the moles of silver nitrate in the solution. We can use the mass percent and density to do this.
First, let's assume we have 100 g of the solution.
This means we have 36 g of silver nitrate in the solution (since it is 36% by mass).
The molar mass of AgNO3 is 169.88 g/mol, so we can convert the mass of the silver nitrate to moles: 36 g / 169.88 g/mol = 0.2124 mol.
The density of the solution is given as 1.41 g/mL, so 100 g of the solution would have a volume of 100 g / 1.41 g/mL = 70.92 mL.
Now, we can calculate the molality using the moles of AgNO3 and the mass of the water:
Molality = moles of solute / mass of solvent (in kg)
Molality = 0.2124 mol / 0.07092 kg
= 2.99 mol/kg
The volume of a single tungsten atom is 1.07×10-23 cm3. What is the volume of a tungsten atom in microliters?
Answer: 1.07×10^-20microlitre
Explanation:
1cm3 = 1000microlitres
1.07×10^-23 cm3 of tungsten = 1.07×10^-23 x 1000 = 1.07×10^-20microlitre
The volume of a single tungsten atom in microliters is 1.07x10^-17 µL. This is found by multiplying the given volume in cubic centimeters by the conversion factor of 1,000,000 µL/cm³.
Explanation:The volume of a single tungsten atom is 1.07×10-23 cm3. One cubic centimeter (cm3) is equal to 1,000,000 microliters (µL). To convert the volume from cubic centimeters to microliters, we need to multiply the original value by the conversion factor. Therefore, the volume of a tungsten atom in microliters will be 1.07×10-23 cm3 * 1,000,000 µL/cm3, which equals to 1.07×10-17 µL. Hence, the volume of a tungsten atom in microliters is 1.07×10-17 µL.
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The human eye is a complex sensing device for visible light. The optic nerve needs a minimum of 2.0 x 10⁻¹⁷ J of energy to trigger a series of impulses that eventually reach the brain.
(a) How many photons of red light (700. nm) are needed?
(b) How many photons of blue light (475 nm)?
Answer:
No. of Photons (red) = 7.05 No. of Photons
No. of Photons (blue) = 4.78 No. of Photons
Explanation:
v (red) = 3 x 108/700 x 10-9 = 4.28 x 10 power 15 Hz
v (blue) = 3 x 108/475 x 10-9 = 6.31 x 10 power 15 Hz
(a) No. of Photons (red) = E/hv = 2.0 x 10 power -17 / 6.626 x 10 power -34 x 4.28 x 10 power 15
No. of Photons (red) = 7.05 No. of Photons
(b) No. of Photons (blue) = E/hv = 2.0 x 10 power -17 / 6.626 x 10 power -34 x 6.31 x 10 power 15
No. of Photons (blue) = 4.78 No. of Photons
Define shielding and effective nuclear charge. What is the connection between the two?
Answer:
The relation between the shielding and effective nuclear charge is given as
[tex]Z_{eff} = Z -S[/tex]
where s denote shielding
z_{eff} denote effective nuclear charge
Z - atomic number
Explanation:
shielding is referred to as the repulsion of an outermost electron to the pull of electron from valence shell. Higher the electron in valence shell higher will be the shielding effects.
Effective nuclear charge is the amount of net positive charge that valence electron has.
The relation between the shielding and the effective nuclear charge is given as
[tex]Z_{eff} = Z -S[/tex]
wheres denote shielding
z_{eff} denote effective nuclear charge
Z - atomic number
The thermite reaction, used for welding iron, is the reaction of Fe3O4 with Al. 8 Al (s) + 3 Fe3O4 (s) ⟶ 4 Al2O3 (s) + 9 Fe (s) Δ H° = -3350. kJ/mol rxn. Because this large amount of heat cannot be rapidly dissipated to the surroundings, the reacting mass may reach temperatures near 3000. °C. How much heat (in kJ) is released by the reaction of 19.3 g of Al with 63.2 g of Fe3O4?
The reaction of 19.3 g of aluminum with 63.2 g of ferric oxide is exothermic, releasing 2395.25 kJ of heat. Aluminum is the limiting reagent in the reaction.
The limiting reagent is the reactant that is used up completely in the reaction. To determine the limiting reagent, compare the ratio of the moles of each reactant to the stoichiometric coefficients in the balanced chemical equation. The reactant with the smaller ratio is the limiting reagent.
1: Calculate the number of moles of each reactant.
Moles of aluminum = 19.3 g / 27 g/mol = 0.715 mol
Moles of ferric oxide = 63.2 g / 231.5 g/mol = 0.272 mol
2: Determine the limiting reagent.
The ratio of moles of aluminum to moles of ferric oxide is 0.715 mol / 0.272 mol = 2.63. The stoichiometric ratio of aluminum to ferric oxide in the balanced chemical equation is 8:3. Since 2.63 is less than 8/3, aluminum is the limiting reagent.
3: Calculate the amount of heat released.
q = -3350 kJ/mol * 0.715 mol = -2395.25 kJ
4: Interpret the answer.
The amount of heat released is -2395.25 kJ. This means that the reaction releases 2395.25 kJ of heat.
The amount of heat released by the reaction of 19.3 g of Al with 63.2 g of Fe_3O_4 is -2395.25 kJ.
when 45.8 g of alanine (C3H7NO2) are dissolved in 1200. g of a certain mystery liquid X, the freezing point of the solution is 4.10 ℃ less than the freezing point of pure X. Calculate the mass of sodium chloride that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor i-1.82 for sodium chloride in X. Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits x n?
Answer:
16.5 g
Explanation:
When a nonvolatile compound is dissolved in a pure solvent, the freezing point of the solvent is reduced, because the interaction solvent-solute requires more energy to be joined, and so, it freezes. This property is called cryoscopy, and the temperature change (ΔT) can be calculated by:
ΔT = Kc*W*i
Where Kc is the cryoscopy constant of the solute X, W is the molality of the solution, and i is the van't Hoff factor, which determines the percent of the solute that is dissolved. For organic molecules, such as alanine, i = 1.
The molality is the number of moles of the solute divided by the mass (in kg) of the solvent (1200 g = 1.2 kg). The molar mass of alanine is 89.09 g/mol, and the number of moles of it is the mass divided by the molar mass:
n = 45.8/89.09
n = 0.5141 mol
W = 0.5141/1.2 = 0.4284 mol/kg
So, Kc of X is:
4.10 = Kc*0.4284*1
Kc = 9.57 °C.kg/mol
So, if now sodium chloride is added to X, and the variation temperature is the same, and i = 1.82:
4.10 = 9.57*W*1.82
W = 0.2354 mol/kg
The number of moles of the solute is then:
W = n/1.2
0.2354 = n/1.2
n = 0.2825 mol
The molar mass of sodium chloride is 58.44 g/mol, thus the mass is the molar mass multiplied by the number of moles:
m = 58.44*0.2825
m = 16.5 g
When the excited electron in a hydrogen atom falls from to , a photon of blue light is emitted. If an excited electron in falls from , which energy level must it fall to so that a similar blue light (as with hydrogen) is emitted? Prove it.
Answer:
n = 3 for similar blue light
Explanation:
The principle applied here is energy levels and energy changes. There are different energy levels depending on the value of the integer as explained by Max planck - a german physicist in 1900, Max planck claimed that electrons in an atom were presumed to be oscillating with a frequency f, then there enrrgy will be given by the plancks equation ; E =hf, where h is the plancks constant.
In general energy of each level can be written as E =nhf
Only certain electron transitions are allowed from one energy level to another. In one-electron species, the change in the quantum number l of an allowed transition must be ±1. For example, a 3p electron can drop directly to a 2s orbital but not to a 2p. Thus, in the UV series, where nfinal = 1, allowed electron transitions can start in a p orbital (l = 1) of n = 2 or higher, not in an s (l = 0) or d (l = 2) orbital of n = 2 or higher. From what orbital do each of the allowed electron transitions start for the first four emission lines in the visible series (nfinal = 2)?
Final answer:
The first four lines of the Balmer series involve electron transitions from 3p to 2s, 4p to 2s, 5p to 2s, and 6p to 2s orbitals.
Explanation:
The Balmer series involves electron transitions from higher energy levels to the second principal energy level (n=2), producing visible spectral lines. For the first four lines of the visible emission spectrum in the Balmer series, the allowed transitions must follow the selection rule Δl = ±1. Therefore, these transitions can only start from orbitals with l=1, which are the p orbitals.
For nfinal = 2, the corresponding ni initial energy levels for the first four visible emission lines are:
3p (n=3, l=1) to 2s (n=2, l=0)4p (n=4, l=1) to 2s (n=2, l=0)5p (n=5, l=1) to 2s (n=2, l=0)6p (n=6, l=1) to 2s (n=2, l=0)How many orbitals in an atom can have each of the following designations: (a) 5f; (b) 4p; (c) 5d; (d) n = 2?
Answer: (a) seven orbitals, (b). 3 orbitals, (c). 5 orbitals and (d). 4 orbitals.
Explanation:
In order to solve this question we need to know how to explain the behaviour of electrons in atoms,and what we need to know is what is called the quantum numbers. There are four different kinds of quantum numbers and they are;
(1). Principal quantum numbers: the principal quantum number is denoted by the letter 'n'. It is used to describe the orbitals' energy. It has the values of n=1,2,3,4,...
(2). The spin quantum numbers: the spin quantum numbers is denoted by m(s). The 's' in the parenthesis is in subscript. It has the values of +1/2 and -1/2.
(3). Azimuthal quantum numbers: this is denoted by ℓ and it is used to explain orbital angular momentum and orbital shape. It has the values of ℓ= 0,1,2,3,....n-1.
Note that => ℓ = 0; we have a s-subshell,sphere shape.
ℓ = 1; p-subshell, dumb bell shape.
ℓ=2; d- subshell, double dumb bell shape.
ℓ= 3; f - subshell, multiple lobes.
(4). Magnetic quantum number: it is denoted by m(l) where the 'l' in the parenthesis is in subscript.
===> NOTE: there are (2ℓ + 1 ) orbitals in a subshell, also, there are n^2 number of orbitals in a shell.
Having known all that above, let us jump right in to the solution.
(a). From above we can see that; there are (2ℓ + 1 ) orbitals' in a subshell, also, f= ℓ= 3.
Therefore, the number of orbitals in 5f = 2 ℓ + 1 = (2×3) + 1 = 6+1 = 7 orbitals for 5f.
(b). 4p, the numbers of orbitals in 4p is; p= ℓ= 1=> 4 ℓ + 1 = (2×1) + 1 = 2+1 = 3 orbitals for 4p.
(c). 5d, the numbers of orbitals' in 5d is; d= ℓ= 2 = (2×2) + 1 = 4 + 1 = 5 orbitals for 5d.
(d). For n= 2, the numbers of orbital is ; n^2. Where the n given is 2. Therefore, 2^2= 2×2 = 4 orbitals in n=2.
An orbital refers to a region in space where electrons can be found.
An orbital refers to a region in space where there is a high probability of finding an electron. Orbitals that posses the same amount of energy are called degenerate orbitals.
The number of orbitals in an atom that can have the following designations are shown below;
5f - seven orbitals can have this designation because the f orbital is seven fold degenerate.4p - three orbitals can have this designation because the p orbital is three fold degenerate5d - five orbitals can have this designation because the d orbital is five fold degeneraten = 2 - the total number of orbitals in an energy level is given by n^2. Hence there are four orbitals that has the designation n =2Learn more: https://brainly.com/question/1527403
Write the full electron configuration of the Period 2 element with the following successive IEs (in kJ/mol):
IE₁ = 801
IE₂ = 2427
IE₃ = 3659
IE₄ = 25,022
IE₅ = 32,822
Answer:
Boron (B) is the element whose IE matches with our data.
Electronic Configuration of boron: [tex]1s^22s^22p^1[/tex]
Explanation:
Ionization Energy (IE):
It is the minimum amount of energy which is required to remove the lose electron. If the electron is closer to the nucleus then greater amount of energy is required to remove the electron.
If we look from left to right in a period, ionization energy increases due stability of valance shell.
From the data given to us:
IE₁ = 801
IE₂ = 2427
IE₃ = 3659
IE₄ = 25,022
IE₅ = 32,822
Boron (B) is the element whose IE matches with our data.
Electronic Configuration of boron: [tex]1s^22s^22p^1[/tex]
Boron has 5 electrons (3 in valance shell) that's why it has 5 Ionization Energies.
Sam records the mass of his evaporating dish as 9.687 g. He records the mass of the evaporating dish and the sample of hydrate as 18.407 g. After heating the sample in the evaporating dish to constant weight, the mass of them combined is 14.007 g. How many moles of water were removed from the sample by the heating process?
Answer:
0.2544 moles of water were removed from the sample by the heating process.
Explanation:
Mass of empty evaporating dish = 9.687 g
Mass of evaporating dish and sample of hydrate = 18.407 g
Mass of hydrate = 18.407 g - 9.687 g = 8.72 g
Mass of the evaporating dish and hydrate after heating = 14.007 g
14.007 g = mass of dish + mass of dehydrate
Mass of dehydrate = 14.007 g - 9.867 g = 4.14 g
Mass of water evaporated = hydrated sample - dehydrated sample
= 8.72 g - 4.14 g = 4.58 g
Moles of water evaporated :
[tex]\frac{4.58 g}{18 g/mol}=0.2544 mol[/tex]
0.2544 moles of water were removed from the sample by the heating process.
The mass of the water removed from the sample by the heating process is determined by subtracting the mass of the evaporating dish and sample after heating from before heating. This mass is then divided by the molar mass of water (18.015 g/mol) to calculate the number of moles of water removed.
Explanation:To calculate the moles of water removed from the sample by the heating process, first we have to determine the mass of the water removed. This would be equal to the mass of the evaporating dish and sample of hydrate before heating minus the mass after heating, i.e., 18.407 g - 14.007 g = 4.4 g.
Then, we calculate the number of moles, using the molar mass of water as a conversion factor from mass to moles. The molar mass of water is 18.015 g/mol. Therefore, the number of moles of water removed would be 4.4 g / 18.015 g/mol = 0.244 mol.
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Draw the partial (valence-level) orbital diagram, and write the symbol, group number, and period number of the element:
(a) [Ar] 4s²3d⁵
(b) [Kr] 5s²4d²
Answer:
a) The element is Manganese (Mn)
b) The element is Zirconium (Zr)
Explanation:
The step by step analysis and explanation is as shown in the attachment
Specify the l and ml values for n = 4.
Answer : The specify the l and ml values for n = 4 are:
At l = 0, [tex]m_l=0[/tex]
At l = 1, [tex]m_l=+1,0,-1[/tex]
At l = 2, [tex]m_l=+2,+1,0,-1,-2[/tex]
At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
Explanation:
There are 4 quantum numbers :
Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of [tex]m_l[/tex] will be -2, -1, 0, +1, +2.
Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.
As we are given, n = 4 then the value of l and ml are,
l = 0, 1, 2, 3
At l = 0, [tex]m_l=0[/tex]
At l = 1, [tex]m_l=+1,0,-1[/tex]
At l = 2, [tex]m_l=+2,+1,0,-1,-2[/tex]
At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
The orange color of carrots and orange peel is due mostly to β-carotene, an organic compound insoluble in water but soluble in benzene and chloroform. Describe an experiment to determine the concentration of β-carotene in the oil from orange peel.
Answer:
First Method: Vacuum Distillation and Chromatographic separation of the remains that were precipitated out from the peel.
Second Method: Extraction of components from orange peels by help of precipitation procedures that are mostly done In Situ. Those components can be recovered using the saponification process. Then these are examined under UV light spectroscopy. Now, the existence and extent of carotenoids can be determined by checking the levels of anti-oxidants.
When measuring water using the graduated pipette, the mass of beaker and the water is 23.670 g. If the empty beaker weighs 13.712 g. What is the volume of water measured assuming a density of 0.9982071 g/mL.
Answer:
The volume of water measured is 10mL
Explanation:
Given;
Mass of mass of beaker and the water = 23.670 g
Mass of empty beaker = 13.712 g
Then, mass of water only = Total mass of of beaker and the water minus Mass of empty beaker
mass of water only = 23.670 g - 13.712 g = 9.958 g
Density = mass/volume
Given density of water = 0.9982071 g/mL
Density of water = Mass of water/ Volume of water
Then, Volume of water = Mass of water/Density of water
Volume of water = 9.958 g/0.9982071 g/mL
Volume of water = 9.975886 mL ≅ 10mL
Therefore, The volume of water measured is 10mL
Hydrogenation reactions, in which H2 and an "unsaturated" organic compound combine, are used in the food, fuel, and polymer industries. In the simplest case, ethene (C2H4) and H2 form ethane (C2H6).
(a) If 138 kJ is given off per mole of C2H4 reacting, how much heat (in kJ) is released when 15.9 kg of C2H6 forms? Enter a positive number since released implies a negative number. Enter to 0 decimal places.
Answer:
73140 kJ.
Explanation:
Equation of the hydrogenation reaction:
C2H4 + H2 --> C2H6
Molar mass = (2*12) + (6*1)
= 30 g/mol
Mass of C2H4 = 15.9 kg
= 15900 g
Number of moles = mass/molar mass
= 15900/30
= 530 mol of C2H6
By stoichiometry, 1 mole of C2H4 is hydrogenated to give 1 mole of C2H6
Number of moles of C2H4 = 530 mole
Q = n * q
= 530 * 138 kJ
= 73140 kJ.
Final answer:
To calculate the total heat released when 15.9 kg of ethane (C₂H₆) is formed, we convert the mass of C₂H₆ to moles and multiply by the heat released per mole of ethene (C₂H₄). The total heat released is 73015 kJ.
Explanation:
The student is asking about the heat released during a hydrogenation reaction where ethene (C₂H₄) and hydrogen gas (H2) are converted into ethane (C₂H₆). The reaction is exothermic, meaning that heat is released when it occurs. Given that 138 kJ is released per mole of C₂H₄ reacting, to find the total heat released when 15.9 kg of C₂H₆ is formed, we need to follow these steps:
First, convert the mass of C₂H₆ to moles using its molar mass (30.07 g/mol for C₂H₆).Since we are starting with the product, C₂H₆, we calculate the moles of C₂H₄ that would have reacted by assuming a 1:1 molar ratio (since each mole of C₂H₆ formed comes from a mole of C₂H₄ reacting).Multiply the moles of C₂H₄ by the heat released per mole to find the total heat released.Here's the calculation:
Mass of C₂H₆ (15.9 kg) = 15,900 gMoles of C₂H₆ = 15,900 g / 30.07 g/mol = 528.95 molHeat released = 528.95 mol × 138 kJ/mol = 73015.1 kJTherefore, the total heat released when 15.9 kg of ethane is formed is 73015 kJ (to zero decimal places).
Question 20 It takes 614./kJmol to break a carbon-carbon double bond. Calculate the maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon. Be sure your answer has the c
The maximum wavelength of light that can break a carbon-carbon double bond by absorbing a single photon is calculated to be 1940 nm. This is done by converting the energy required to break the bond to J/particle, and then using this to find the wavelength using the equation E=h*c/λ.
Explanation:To find the maximum wavelength of light that can break a carbon-carbon double bond by absorbing a single photon, it is necessary to convert the energy required to break the bond from kJ/mol to energy per photon and then use that to calculate the wavelength. Using the relation between energy and wavelength given by the formula E=h×c/λ, where E is Energy, h is Planck's constant (6.626 x 10⁻³⁴ Js), c is the speed of light and λ is the wavelength.
First, convert the energy required to break the bond to J/particle by converting kJ to J (1kJ = 1000J) and then dividing by Avogadro's number (6.022 x 10²³). Thus, E = 614.4 kJ/mol × 1000 J/kJ / 6.022 x 10²³ particles/mol = 1.02 x 10⁻¹⁹ J/particle.
Then, rearrange the formula to solve for λ. We get λ = h × c/E. Substituting values (h = 6.626 x 10⁻³⁴ Js, c = 3.00 x 10⁸ m/s, and E = 1.02 x 10⁻¹⁹ J) gives λ = 1.94 x 10⁻⁶ meters or 1940 nm.
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The maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon is 341 nm.
Explanation:To calculate the maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon, we need to use the formula E = hc/λ, where E is the energy of the photons, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength of light in meters.
First, we need to convert the bond energy from kJ/mol to J/molecule by multiplying it by Avogadro's number (6.02 x 10^23). So, the bond energy is 614 x 10^3 J/mol. Next, we can rearrange the formula to solve for λ:
λ = hc/E = (6.63 x 10^-34 J·s)(3.0 x 10^8 m/s)/(614 x 10^3 J/mol)
Calculating this expression gives us a value of λ = 3.41 x 10^-7 m, or 341 nm. Therefore, the maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon is 341 nm.
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Use valence bond theory to describe the bonding in hydrogen sulfide (H2S). What are the predicted bond angles in hydrogen sulfide?
Answer:
The bond angle between the p orbital is assumed to be 90 degrees But according to VSEPR theory, as molecular geometry is a little bit bent thus the angle is less than 109.5 degrees.
Explanation:
IN hydrogen sulfide, H_2 S, both hydrogen atoms consist of para-magnetic 1s orbital while sulfur consists of diamagnetic 3s and 2 para magnetic 3p orbital.
The bond angle between the p orbital is assumed to be 90 degrees But according to VSEPR theory, as molecular geometry is a little bit bent thus the angle is less than 109.5 degrees.
In hydrogen sulfide (H2S), the bonding is explained using valence bond theory. The bond angle in H2S is approximately 92 degrees.
Explanation:In hydrogen sulfide (H2S), the bonding can be explained using valence bond theory. According to this theory, covalent bonds are formed by the overlap of atomic orbitals. In H2S, the sulfur atom forms a covalent bond with each hydrogen atom through overlap of its sp3 hybrid orbital and the hydrogen 1s orbital.
The predicted bond angle in H2S is approximately 92 degrees. This angle is less than the ideal tetrahedral angle of 109.5 degrees due to the presence of two lone pairs of electrons on the sulfur atom, which exert greater repulsion than the bonding pairs. This results in a distorted tetrahedral shape with a smaller bond angle.
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Prior to hosting an international soccer match, the local soccer club needs to replace the artifical turf on their field with grass turf. The grass turf will cost $ 9.75 per square meter. If the field is 0.102 km by 0.069 km, how much will it cost the club to add the grass turf to their field?
It will cost the soccer club $68,618.50 to add the grass turf to their field for the international soccer match.
Given: Length = 0.102 km = 0.102 km × 1000 m/km = 102 m
Width = 0.069 km = 0.069 km × 1000 m/km = 69 m
The area of the field: Area = Length × Width
Area = 102 m × 69 m = 7038 sq. meters
The area by the cost of the grass turf per square meter:
Cost = Area × Cost per square meter
Cost = 7038 sq. meters ×$9.75/sq. meter = $68,618.50
Therefore, it will cost the soccer club $68,618.50 to add the grass turf to their field for the international soccer match.
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